Answer:
392 N
Explanation:
Draw a free body diagram of the rod. There are four forces acting on the rod:
At the wall, you have horizontal and vertical reaction forces, Rx and Ry.
At the other end of the rod (point X), you have the weight of the sign pointing down, mg.
Also at point X, you have the tension in the wire, T, pulling at an angle θ from the -x axis.
Sum of the moments at the wall:
∑τ = Iα
(T sin θ) L − (mg) L = 0
T sin θ − mg = 0
T = mg / sin θ
Given m = 20 kg and θ = 30.0°:
T = (20 kg) (9.8 m/s²) / (sin 30.0°)
T = 392 N
I bet it is too bad to get in it but I don’t want it in my anymore cause it’s a good boy to be mad cause he doesn’t even care anymore anymore lol
To balance an equation, you do whatever you did on one side, on the other side. If you subtract, then you subtract on the other side. If you multiply, you multiply on the other side, etc...
Allowing for a platform in which the box holds potential energy
